- #1
jostpuur
- 2,112
- 19
I have some reasons to believe that this equation is true:
<br /> \lim_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{(n!)^2} = \frac{1}{\sqrt{\pi}}<br />
Anyone having idea of the proof? I don't even know how to prove that the limit is strictly between zero and infinity.
<br /> \lim_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{(n!)^2} = \frac{1}{\sqrt{\pi}}<br />
Anyone having idea of the proof? I don't even know how to prove that the limit is strictly between zero and infinity.
